Proof of the Irrational

Numbers, that is.

The mathematics of infamous Ancient Greeks was based on integers and rational numbers.¹ The latter are comprised of two integers, so all their calculations were essentially integer-based. The rational numbers — in the form a/b — can be as small or as large as required. And there is always another rational number between any two rational numbers, no matter how close together those are. They seem to cover the number line, so the Greeks thought the rational numbers were all that was needed.

A cornerstone of their math² is the Pythagorean theorem:

\(\textsf{Hypotenuse}=\sqrt{{a}^{2}+{b}^{2}}\)

Which relates the lengths of the two short sides (a and b) to the length of the hypotenuse of a right triangle.

This fundamental theorem turned out to be a huge troublemaker for the Pythagoreans because of the simple case where the two short sides both have a length of one. A common example might be the diagonal of a square tile with all four sides one foot in length.

The problem is what the Pythagorean theorem says for this:

\(\textsf{Diagonal}=\sqrt{ {1}^{2} \,+\, {1}^{2}\;}=\sqrt{2}\)

The diagonal length across a square tile (with one-foot sides) is the square root of two. The Greeks knew about square roots — those go way back. Obviously, the number needed isn’t an integer. We need some number whose square is two. The Greeks assumed it was some fraction:

\(\left(\frac{a}{b}\right)^{2}=\frac{a^2}{b^2}={2}\)

You can try searching for values of a and b that solve the equation, but no matter how long you search, you’ll never find integer values of a and b that give the desired result (of exactly two).

The Greeks eventually proved that no values for a and b existed, which was a big — and very troubling — development because it meant the rational numbers weren’t enough. There were (as it turns out) lots (and lots) of numbers that weren’t rational, they were irrational. This was a hard pill for the Ancient Greeks to swallow.

With that history as context, recently I read a cute proof demonstrating that no integer solutions exist for a and b above.

That proof goes like this…

Start by assuming some integer a and b exist such that:

\(\frac{a^2}{b^2}={2}\)

Now multiply both sides by b²:

\({a}^{2}={2}{b}^{2}\)

This tells us that:

\({0}\lt{b}^{2}\lt{a}^{2}\)

Because we know a² is two times b².

And since a² is even (it’s twice b², so it has to be), a must be even (because odd numbers have odd squares). So now let’s create a new value, c, such that:

\(\frac{a}{2}={c}\;\;\rightarrow\;\;{a}={2}{c}\)

So now, in the equation above, we can replace a with 2c (and bring the 2b² to the front):

\({2}{b}^{2}=\left({2c}\right)^{2}={4}{c}^{2}\)

Now divide both ends by two:

\({b}^{2}={2}{c}^{2}\)

As above, this tells us that:

\({0}\lt{c}^{2}\lt{b}^{2}\lt{a}^{2}\)

And the new equation has the same pattern we started with, so we can repeat the process with b that we did above with a:

\(\frac{b}{2}={d}\;\;\rightarrow\;\;{b}={2}{d}\)

And then:

\({2}{c}^{2}={4}{d}^{2}\;\;\rightarrow\;\;{c}^{2}={2}{d}^{2}\)

Giving us again the same pattern as well as telling us that:

\({0}\lt{d}^{2}\lt{c}^{2}\lt{b}^{2}\lt{a}^{2}\)

It should be obvious this can’t continue indefinitely because each squared value has to be a positive integer (i.e. greater than zero). The bottom line is that assuming a and b exist leads to a contradiction, so the assumption must be false. There are no values for a²/b² that equals two.

A way to see more clearly that the progression can’t continue is to rewrite the less-than chain in terms of the last variable we defined, d:

\({0}\lt{d}^{2}\lt{2}{d}^{2}\lt{4}{d}^{2}\lt{8}{d}^{2}\)

Which we can write (substituting x for d) as:

\({0}\lt{2}^{0}{x}^{2}\lt{2}^{1}{x}^{2}\lt{2}^{2}{x}^{2}\lt{2}^{3}{x}^{2}\)

If we divide each term by the last factor, we get:

\({0}\lt{2}^{-3}{x}^{2}\lt{2}^{-2}{x}^{2}\lt{2}^{-1}{x}^{2}\lt{2}^{0}{x}^{2}\)

Which is:

\({0}\lt\frac{{x}^{2}}{8}\lt\frac{{x}^{2}}{4}\lt\frac{{x}^{2}}{2}\lt{x}^{2}\)

Making it very clear the sequence cannot continue. At some point, x² cannot be divided by a factor of two with an integer result.

That’s all for this time. I thought it was a clever way to prove there is no rational solution to the diagonal of a square.³ And as we now know:

\(\sqrt{2}={1.4142135623730950488016887242097}\ldots\)

Which, by the way, means a rough rational approximation is:

\(\sqrt{2}\approx\frac{1414213}{1000000}={1.414213}\)

Accurate to six decimal places! And it’s fairly easy to remember “one, forty-one, forty-two, thirteen”.⁴

Until next time…

1. And geometry. The A.G. were huge on geometry.

2. And geometry.

3. Any square of any size because any square with sides different from unity simply scales up (or down) its dimensions, including the diagonal, so the diagonal is always a factor of √2.

4. Science fiction fans can remember it as “one, the number before forty-two, forty-two, thirteen”, the last number being the base in which “what do you get if you multiply six by nine?” the answer forty-two is true.